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Use the question's decreasing survivor function convention $F(t)=\mathbb{P}(T>t)$, rather than interpreting $F$ as a cumulative distribution function. An exact failure at $t$ contributes the mass $F(t-)-F(t)$. The right censoring contribution is $F(3)$, and the interval censoring contribution is $F(2)-F(5)$. Under independent observations and a noninformative observation mechanism, the empirical likelihood with mixed censoring, omitting observation-mechanism factors, isThis is a likelihood over probability distributions with possible atoms; replacing the exact-event masses by a smooth density would be a different model.

Set $u=F(2)$ and $v=F(5)$. Monotonicity and $0\le F\le 1$ giveFor instance, $F(4-)\le F(2)=u$ while $F(4)\ge F(5)=v$, proving the second inequality. The interval-censoring factor already equals $u-v$. Multiplying gives the monotonicity bound for a mixed-censoring likelihoodFor any $0<v<u<1$, this bound is attained by a distribution with masses $1-u$, $u-v$ and $v$ at $1$, $4$ and $6$ respectively. Its survivor function is one before the first mass, $u$ between the first and second, $v$ between the second and third, and zero thereafter.

The maximum likelihood is positive, because such a three-mass distribution has positive likelihood. At any maximizer all the above inequalities must therefore be equalities; otherwise replacing it by that three-mass distribution at the same $u,v$ would strictly improve its likelihood. Equality, together with monotonicity, yieldsThus these relations are consequences of maximum likelihood, not assumptions about where an arbitrary distribution must put its mass.

The reduced log-likelihood isIts stationary equations areThe second gives $u=3v$. Substituting in the first gives $-1/(1-u)+4/u=0$, henceThe reduced log-likelihood is strictly concave: its second directional derivative in a nonzero direction $(a,b)$ isThe likelihood vanishes on the boundary of the admissible triangle, so this stationary point is the unique maximum. The corresponding nonparametric maximum-likelihood estimator puts masses $1/5$, $8/15$ and $4/15$ at the three exact failure times.

Condition on the claim count distribution. If $\mu =\mathbb{E}{X}_1$ and $v=\mathrm{Var}(X_1)$, then independence of the count and the independent and identically distributed random variables givesThe empty sum is zero. The law of total expectation and law of total variance therefore prove the random sum of independent claims formulasFor the variance calculation assume the count and severity have finite second moments; the expectation identity needs the corresponding integrability. Nonnegative claims also permit the expectation identity with extended nonnegative values.

Represent each policy payment as $S_i=B_i{Y}_i$, where $B_i$ is a Bernoulli random variable of parameter $q_i$, independent of its severity. Equivalently, $Y_i$ specifies the payment distribution conditional on a claim. ThenIndependence of the policies makes their variances add. PutThe heterogeneous individual claims model consequently has

For the Poisson approximation, define the severity mixture distributionLet ${\phi }_i(t)=\mathbb{E}[e^{it{Y}_i}]$ be the individual characteristic functions. Conditioning on each Poisson count and then multiplying transforms of the independent policy payments givesThis is Poisson superposition of insurance portfolios. By the uniqueness theorem for characteristic functions,with count and severities independent. Its compound Poisson distribution has distribution functionwhere $F^{\ast k}$ is the convolution distribution of $k$ independent severities and $F^{\ast 0}$ is the distribution concentrated at zero. Characteristic functions avoid assuming that severity moment-generating functions exist. Since $\mathbb{E}Z=M/Q$ and $\mathbb{E}{Z}^2=A/Q$, the random-sum formulas give

Use $T_{\mathrm{B}}$ to distinguish the second, compound-binomial approximation from the Poisson one. It has $K_{\mathrm{B}}\sim \mathrm{Bin}(n,p)$ with $p=Q/n$ and the same severity law $F$. Thus $\mathbb{E}{K}_{\mathrm{B}}=Q$ and $\mathrm{Var}(K_{\mathrm{B}})=Q(1-p)$. Applying the random sum of independent claims identities again givesHence

Both approximations match the exact expected value. The Cauchy-Schwarz inequality gives $M^2\le nC$, proving the variance comparison for compound portfolio approximations:The excess variances are respectively $C-M^2/n$ and $C$. The compound-binomial model is therefore closer in variance; equality with the exact variance occurs when all $q_i{\mu }_i$ are equal. This variance comparison is not a general ordering of approximation errors or tail probabilities.

The Poisson replacement is useful when all claim probabilities are small: it replaces the individual restriction of at most one claim by an unrestricted Poisson count, and the variance error is of second order in those probabilities. The compound-binomial replacement retains a count bounded by $n$, but pools both frequencies and severity laws and can lose policy-specific information. If all $q_i$ and all $F_i$ agree, it is exactly the original portfolio law. In a heterogeneous portfolio, agreement of means, or even variances, does not make either approximate law exact. Neither approximation is guaranteed accurate merely by having many policies.

The logit link makes the probability constant within each group: write $p_i=\{1+\exp [-(\mu +{\alpha }_i)]{\}}^{-1}$, with ${\alpha }_1=0$. Let $S_i={\sum }_{j=1}^k{Y}_{ij}$ and $N=km$. Independence gives $S_i\sim \mathrm{Bin}(N,p_i)$. Up to terms independent of the parameters, the log likelihood isThe likelihood equations areTogether they also imply $S_1-N{\hat{p}}_1=0$. Thus, when all totals satisfy $0<S_i<N$,The group logits are unconstrained apart from their parametrization, and each binomial log likelihood is strictly concave in its logit, so these equations give the unique finite maximum. A total of $0$ or $N$ instead gives a boundary probability and an infinite logit; a finite MLE need not exist in that case.

For $n=2$, keep $0<p_1,p_2<1$ fixed and let $N=km\to \infty$. Since ${\hat{p}}_1=S_1/N$, the central limit theorem gives $\sqrt{N}({\hat{p}}_1-p_1)\stackrel{d}{\to }N(0,p_1(1-p_1))$. The derivative of the logit is $1/[p_1(1-p_1)]$, so the delta method gives the baseline logit estimator in a group-factor binomial model lawThe second group does not improve the baseline estimate because it has its own free offset. Equivalently, the Fisher information matrix for $(\mu ,{\alpha }_2)$ isIts first diagonal entry confirms the same asymptotic variance.

We prove the positive state-price density alternative using finite-dimensional convex cones. Put $w=(1+\Vert A_1\Vert {)}^{-1}$ and defineAll displayed expectations are finite, since $w\Vert A_1\Vert \le 1$. The set $D$ is a convex cone, so $K$ is a closed convex cone. A vector $x$ is nonnegative on $K$ exactly when $x^{\top }{A}_1\ge 0$ almost surely. Indeed, one direction follows by taking every nonnegative bounded test weight $f$; for the other direction, if $x^{\top }{A}_1<0$ on a set of positive probability, take its indicator as $f$ to obtain a strictly negative expectation. This also identifies $K$ as the closed cone generated by the support of a probability distribution of $A_1$.

The three cases $A_0\notin K$, $A_0\in K\setminus \mathrm{ri}K$, and $A_0\in \mathrm{ri}K$ are treated below. They give respectively a negative-cost separating portfolio, a supporting zero-cost portfolio with a strictly positive payoff on an event of positive probability, or the required positive pricing weight. Thus the alternatives are exhaustive.

They are mutually exclusive. If $A_0=\mathbb{E}[\nu A_1]$ with $\nu >0$ and a portfolio has nonnegative payoff, then $x^{\top }{A}_0=\mathbb{E}[\nu x^{\top }{A}_1]\ge 0$. A negative initial cost is impossible. If the payoff is strictly positive with positive probability, this expectation is strictly positive, so nonpositive initial cost is impossible too. The given weighted integrability permits taking the expectation of this scalar product.

In a one-period market, $x^{\top }{A}_0$ is the portfolio's initial cost and $x^{\top }{A}_1$ its terminal value. The first two cases are investment-consumption arbitrage: nonnegative wealth can be taken out at both dates, with a strict benefit at one of them. If a strictly positive bank account is traded, any negative initial cost can be invested in it, converting the opportunity into a zero-cost arbitrage with nonnegative terminal payoff and positive payoff with positive probability.

In the third case, $\nu$ is a strictly positive state-price density. If a traded numéraire has price $B_0>0$ and terminal value $B_1>0$, its pricing identity is $\mathbb{E}[\nu B_1]=B_0$. ConsequentlyThe density has mean one and is strictly positive, so $Q$ is equivalent to the physical measure. Thus discounted asset prices are martingales in the one-period sense. This is the fundamental theorem of asset pricing formulation of absence of arbitrage; normalization to a probability measure uses the positive traded numéraire, not an unsupported assumption that $\mathbb{E}\nu =1$.

The resulting revenue comparison for two uniform private values isThus the optimally reserved ascending auction is best, followed by the optimal posted price, followed by the ascending auction without a reserve. These values are in the normalized monetary units used for the valuations.

The revenue equivalence theorem for symmetric independent private values applies to mechanisms with the same allocation probabilities at every type and the same expected utility for the lowest type, under the usual incentive and risk-neutrality assumptions. The reserved and unreserved auctions have different allocation rules. Without a reserve, a bidder of valuation $v$ wins with probability $G(v)=v$. With reserve $r$, that probability is zero for $v<r$ and $v$ for $v>r$. In particular the car is withheld when even the highest valuation is below $r$.

The interim payment identity makes the difference explicit. With lowest-type utility zero, the ordinary auction has expected payment $m(v)=v^2/2$. The reserved auction hasSumming the two bidders' ex ante payments gives $2{\int }_r^1{m}_r(v)dv=1/3+r^2-4r^3/3$, exactly the revenue calculated above. Hence there is no conflict with revenue equivalence: its equal-allocation hypothesis does not hold.

In a transferable utility game, the characteristic function of a coalitional game assigns each coalition $S$ the total payoff $v(S)$ that its members can attain or guarantee together, normalized by $v(\varnothing )=0$. If disjoint coalitions $S,T$ can combine their separate strategies without interference, their union can secure both payoffs, givingThis is a superadditive coalitional game. The pooling assumption explains the intended assertion, but superadditivity is not automatic for an abstract characteristic function: the two-player values $v(\{1\})=v(\{2\})=v(\{1,2\})=1$, with empty value zero, are a counterexample. Thus the word “always” requires either that pooling assumption or a definition restricting the class of games to superadditive ones.

For the given bankruptcy game, the total deficit is $19-15=4$, so the coalition value is $({\sum }_{i\in S}{c}_i-4{)}_{+}$. Its computed values areIt is indeed superadditive: for disjoint claim sums $p,q\ge 0$, $(p+q-4{)}_{+}\ge (p-4{)}_{+}+(q-4{)}_{+}$. If both right-hand terms are positive the difference is $4$; if only one is positive, adding the other claim sum cannot reduce the value; if neither is positive the right side is zero.

An almost complex structure is a smooth real vector bundle endomorphism $J:TM\to TM$ satisfying $J^2=-I$. In particular, $M$ has even real dimension $2n$. Extend $J$ complex-linearly to $T_{\mathbb{C}}M=TM{\otimes }_{\mathbb{R}}\mathbb{C}$. Its minimal polynomial divides $(u-i)(u+i)$, whose roots are distinct, so the type decomposition of the complexified tangent bundle isThe smooth projections are $(I-iJ)/2$ and $(I+iJ)/2$. Complex conjugation exchanges the two rank-$n$ eigenbundles. A $(1,0)$-covector annihilates $T^{0,1}M$ and a $(0,1)$-covector annihilates $T^{1,0}M$. Thus the differential forms of type (p, q) are the smooth sections ofHere the tensor product is identified with its image under the wedge product of differential forms. For a real tangent vector $v$, the vectors $v-iJv$ and $v+iJv$ exhibit the two types directly.

We now construct the almost complex structure from a decomposable volume form. Locally write $\Omega ={\theta }_1\wedge \cdots \wedge {\theta }_n$. The nonvanishing of $\Omega \wedge \overline{\Omega }$ says that the $2n$ covectors ${\theta }_1,\dots ,{\theta }_n,{\overline{\theta }}_1,\dots ,{\overline{\theta }}_n$ form a complex coframe. In particular, the span of the ${\theta }_j$ has the intrinsic descriptionIndeed, expansion in that coframe shows that every coefficient of a conjugate factor must vanish if $\alpha \wedge {\Omega }_x=0$. Therefore this span is independent of the chosen local factorization. The local coframes show that $W$ is a smooth rank-$n$ vector subbundle, andDefine $J_{\Omega }^{\ast }$ to be multiplication by $i$ on $W$ and by $-i$ on $\overline{W}$. It commutes with complex conjugation, so is the complexification of a real smooth bundle endomorphism, whose dual defines $J_{\Omega }$ on $TM$. It has square $-I$, and its $(1,0)$-covectors are exactly $W$. Conversely, any such almost complex structure must have this same action on the entire coframe. Thus $J_{\Omega }$ exists globally and is unique, and every factor in every allowed decomposition is a $(1,0)$-form.

Finally suppose $d\Omega =0$. Since ${\theta }_j\wedge \Omega =0$, the exterior derivative product rule givesDecompose $d{\theta }_j$ into types $(2,0)$, $(1,1)$ and $(0,2)$ using $J_{\Omega }$. The first two wedge products with the $(n,0)$-form $\Omega$ vanish by dimension. On $(0,2)$-forms, wedging with $\Omega$ is injective: the monomials $\Omega \wedge {\overline{\theta }}_a\wedge {\overline{\theta }}_b$ are linearly independent. For $n=1$ the $(0,2)$ space is zero, so the statement still holds. Hence every $d{\theta }_j$ has no $(0,2)$ component. For an arbitrary $(1,0)$-form $\alpha ={\sum }_j{a}_j{\theta }_j$, the extra terms $d{a}_j\wedge {\theta }_j$ likewise have only types $(2,0)$ and $(1,1)$. The bracket and differential-form criteria for integrability proved below now giveLocal decomposability has been used essentially; the nonvanishing top-degree product alone does not provide the required coframe.

Write $[M{]}^{\omega }$ for the set of all infinite subsets of $M$. A family $Y\subseteq [\mathbb{N}{]}^{\omega }$ is a Ramsey family in the homogeneous-cone sense if there is an infinite $M$ such that either $[M{]}^{\omega }\subseteq Y$ or $[M{]}^{\omega }\cap Y=\varnothing$. Thus membership in $Y$ is constant on an infinite homogeneous cone. We will prove the stronger version for open families: such a cone exists inside every prescribed infinite ground set, as in the definition of a Ramsey set of infinite subsets.

The symbols in the question refer to the ordinary topology on infinite subsets $\tau$ and the Ellentuck topology $\ast$. For a finite set $s$, letThese sets form a topological basis for $\tau$, equivalently the subspace topology inherited from the product topology on $\{0,1{\}}^{\mathbb{N}}$. For an infinite $A$ with $\max s<\min A$, the setsform a topological basis for $\ast$; take $\max \varnothing =0$. In particular $[s]=[s,\{n:n>\max s\}]$, so $\ast$ is finer than $\tau$.

For a non-Ramsey example use the finite-symmetric-difference parity colouring. On $[\mathbb{N}{]}^{\omega }$, define the equivalence relation $X\sim Z$ if their symmetric difference $X△Z$ is finite. By the axiom of choice, select a representative $R$ in every equivalence class. Give $X$ the colourwhere $R_X$ is its class representative. Deleting one point stays in the same class and toggles the parity. Thus for every infinite $M$, the sets $M$ and $M\setminus \{\min M\}$ are both in $[M{]}^{\omega }$ but have opposite colours. Either colour class, regarded as a family $Y$, is therefore not Ramsey.

Now let $U$ be $\tau$-open. We prove that for every infinite $A_0$ there is an infinite $C\subseteq A_0$ with $[C{]}^{\omega }\subseteq U$ or $[C{]}^{\omega }\cap U=\varnothing$. The proof uses acceptance and rejection of finite stems, which we define explicitly. For a finite stem $s$ and an infinite tail $A$ above $\max s$, say that $A$ accepts $s$ if $[s,A]\subseteq U$, and rejects $s$ if no infinite subset of $A$ accepts $s$. Both properties persist on taking infinite subsets. Every tail has an infinite subset that decides $s$: take an accepting subset if one exists, and otherwise it already rejects $s$.

First thin $A_0$ to decide the empty stem. Then construct $b_1<b_2<\cdots$ and nested infinite tails. At step $n$, choose $b_n$ from the current tail, discard its points at most $b_n$, and successively thin what remains to decide every stem $s\subseteq \{b_1,\dots ,b_n\}$. There are only finitely many such stems, so an infinite tail remains. Let $B=\{b_1,b_2,\dots \}$ and write $B/s=\{b\in B:b>\max s\}$. For every finite $s\subseteq B$, the tail $B/s$ decides $s$, since it lies in the tail obtained at the stage when the largest point of $s$ was selected. The empty stem is also decided. This is deciding all finite stems by fusion.

If $B$ accepts the empty stem, then $[B{]}^{\omega }\subseteq U$, as required. Otherwise $B$ rejects it. We need the finitely many accepting extensions of a rejected stem observation. Suppose a tail $B/s$ rejects $s$. Only finitely many $a\in B/s$ can have $B/a$ accepting $s\cup \{a\}$. Indeed, if infinitely many such $a$ formed a set $D$, every member of $[s,D]$ would have some first new point $a$ and all later points in $B/a$, hence would lie in $U$. This would make $D$ an accepting subset of $B/s$, contradicting rejection.

Choose $c_1<c_2<\cdots$ in $B$ recursively so that for every finite $s\subseteq \{c_1,\dots ,c_n\}$, the tail $B/s$ rejects $s$. The empty stem satisfies this initially. At the next step, each of the finitely many previous stems has only finitely many accepting one-point extensions by the observation. Choose $c_{n+1}$ larger than all previous points and outside all those finite exceptional sets. The first fusion ensured that the tail for every new stem decides it; since it does not accept it, it rejects it. This maintains the recursion.

Put $C=\{c_1,c_2,\dots \}$. If some $X\in [C{]}^{\omega }$ belonged to $U$, ordinary openness would give a finite initial segment $s$ of $X$ with $[s]\subseteq U$. In particular $[s,C/s]\subseteq U$. But $C/s$ is an infinite subset of $B/s$, which rejects $s$, a contradiction. Hence $[C{]}^{\omega }\cap U=\varnothing$. We have proved

The Blaschke product here is taken over distinct points of the orbit; its existence is part of the hypothesis. It is not necessary to assert that every discrete orbit satisfies the Blaschke condition.

The modulus of a Blaschke factor is the pseudohyperbolic distance. Thus, away from the orbit,The logarithms converge absolutely at such a point, by the compact convergence estimate for the Blaschke product. Every $T\in G$ preserves the hyperbolic metric and permutes the orbit. Reindexing this convergent product yields $|B(Tz)|=|B(z)|$.

The functions $B\circ T$ and $B$ have the same simple zero set, since $T$ is a biholomorphism. Their quotient therefore extends holomorphically and without zeros across the orbit. It has modulus one everywhere. The open mapping theorem makes it a constant, denoted $\chi (T)$. FinallySince $B$ is not identically zero, cancellation givesThis is the automorphy character of an orbit Blaschke product. The equal-modulus argument is essential: equality of zero sets alone would not make the quotient of two bounded holomorphic functions constant.

A point fixed by $F^4$ satisfies $(2^4-1)x\in \mathbb{Z}$, so $x=j/15$ for $0\le j\le 14$. The points with period dividing two are $0,1/3,2/3$, corresponding to $j=0,5,10$. All other twelve points have least period four. Multiplication by two modulo fifteen groups them into the three cyclesThis exhausts the requested four-cycles. The displayed order is the dynamical order, and cyclic changes of the starting point describe the same cycle.

Reduce the polynomial modulo $p$, obtaining a nonzero degree-$d$ polynomial over the field ${\mathbb{F}}_p$. If $a$ is a root, polynomial division gives $\bar{f}(X)=(X-a)h(X)$. Any different root $b$ is a root of $h$, because $b-a\ne 0$ is invertible. Induction on degree therefore proves there are at most $d$ distinct roots modulo $p$.

Every nonzero residue is a root of $X^{p-1}-1$ by Fermat's little theorem. The difference between this monic polynomial and the product over all its $p-1$ nonzero roots has degree at most $p-2$, yet has $p-1$ roots. The root bound forces it to vanish identically over ${\mathbb{F}}_p$. Thus all coefficients of the corresponding integer polynomial are divisible by $p$.

For the requested failure modulo $p^2$, take $p=7$: $6!+1=721=7\cdot 103$, and $103$ is not divisible by seven. Hence the prime-square strengthening fails, even though some special primes do satisfy it.

The following unheaded requests concern the spherical measure, the wave solution and both uniqueness arguments. The missing part (i) above and this root Solution slot restore their coverage while retaining the existing part (ii) anchor.

For $k=|\xi |$, rotational symmetry of the spherical measure giveswith continuous value $4\pi t^2$ at $k=0$. The spatial Fourier transform of the wave equation gives ${\hat{u}}_{tt}+k^2\hat{u}=0$ with initial data zero and $\hat{g}$. Thus $\hat{u}=\sin (tk)\hat{g}/k$. The convolution theorem gives the Kirchhoff formulaIt is smooth for Schwartz data, solves the transformed equation, and has $u(0,x)=0$, $u_t(0,x)=g(x)$ by the second expression and its small-time limit.

To prove uniqueness for arbitrary $C^2$ solutions, take their difference $v$, with zero initial data, and set $e=(v_t^2+|\nabla v{|}^2)/2$, $p=-v_t\nabla v$. Direct differentiation gives $e_t+\nabla \cdot p=v_t(v_{tt}-\Delta v)=0$. For any centre $x_{\ast }$ and radius $R$, integrate over the shrinking ball $B(x_{\ast },R-t)$. Differentiating the moving-domain energy givessince $v_t{\partial }_{n}v\le (v_t^2+|\nabla v{|}^2)/2$. Initial energy is zero and energy is nonnegative, so it stays zero. In each such cone $v_t$ and $\nabla v$ vanish, and the initial value is zero, hence $v=0$. Choosing $R>T$ and arbitrary centre proves uniqueness at any time $T$, without imposing finite total energy or decay on the competing solution.

For the equation with the smooth positive time-independent potential $V$, use insteadThen ${\partial }_t{e}_V+\nabla \cdot p=v_t(v_{tt}-\Delta v+Vv)=0$. The same shrinking-ball estimate is nonpositive because $V{v}^2\ge 0$. Zero initial data therefore give $v=0$. The potential-wave initial-value problem has at most one $C^2$ solution, with no global energy-growth assumption required. Time independence of $V$ is what makes this particular local energy identity exact.

Communication is reflexive because $(P^0{)}_{ii}=1$, and symmetric by its definition. If $i\leftrightarrow j$ and $j\leftrightarrow k$, choose positive-probability paths $i\to j$ in $m$ steps and $j\to k$ in $n$ steps. The Chapman-Kolmogorov equation givesCombining the reverse paths gives a positive-probability path $k\to i$ as well. Thus communication is transitive and is an equivalence relation.

To transfer recurrence, first establish the recurrence criterion by return probabilities. Put $r_i={\mathbb{P}}_i(T_i^{+}<\infty )$ and let $N_i={\sum }_{n\ge 0}{1}_{\{X_n=i\}}$ under ${\mathbb{P}}_i$. After each return the Strong Markov property restarts the chain at $i$, so ${\mathbb{P}}_i(N_i\ge k+1)=r_i^k$. ThereforeFor $r_i<1$ this equals $(1-r_i{)}^{-1}$; for $r_i=1$ it is infinite. The expectation identity follows by summing the nonnegative indicators, so remains valid when the expectation is infinite. Consequently $i$ is a recurrent state exactly when its return-probability series diverges.

If $i\leftrightarrow j$, choose $a,b$ with $(P^a{)}_{ji}>0$ and $(P^b{)}_{ij}>0$. For every $n\ge 0$,Summing over $n$, recurrence of $i$ forces the return-probability series for $j$ to diverge. The proved criterion shows $j$ is recurrent, so recurrence is a property of the communicating class.

The sequence has uniform convergence to $f$ on $(0,1)$ precisely whenThe same $N$ must work throughout the interval; equivalently ${\sup }_{0<x<1}|f_n(x)-f(x)|\to 0$.

A real-valued function $f$ on an interval $I$ is uniformly continuous whenThe essential point in uniform continuity is that $\delta$ is independent of the locations of $x$ and $y$. Uniform continuity on the whole real line does not imply boundedness: $f(x)=x$ is Lipschitz continuous with constant $1$, hence uniformly continuous, but is unbounded.

For a steady inviscid incompressible fluid of constant density in uniform gravity $\mathbf{g}=-g{\mathbf{e}}_z$, the Euler equation for fluid motion and continuity equation areTaking the scalar product with $\mathbf{u}$ givesThus Bernoulli's quantity is constant along each streamline. For an irrotational connected flow it is constant throughout the flow; constancy across different streamlines is not assumed in general.

Since incompressibility makes $\nabla \cdot (\rho \mathbf{u}\otimes \mathbf{u})=\rho (\mathbf{u}\cdot \nabla )\mathbf{u}$, integrating over a fixed control volume and applying the divergence theorem yields the integral momentum balanceHere $\mathbf{n}$ is outward and the left side is the net outgoing momentum flux. Uniform atmospheric pressure has zero resultant on a closed control surface, so it may be subtracted from all pressures in the jet calculations.